The equation for the area of a right triangle is:

 \(\displaystyle A=\frac{1}{2}b\times h\)

In this case, our values are: \(\displaystyle b=x, h=(x+1)\)

Plugging this into the equation leaves us with:

\(\displaystyle A=\frac{1}{2}(x)(x+1)\)

which can be rewritten as \(\displaystyle A=\frac{x}{2}(x+1)\)

In order to calculate the area of a right triangle, we need to know the height and the base. We are only given the length of the base, so we first need to use the Pythagorean theorem with the given hypotenuse to find the length of the height:

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle a^2=c^2-b^2=13^2-12^2=169-144\)

\(\displaystyle a^2=25\rightarrow a=5\)

Now that we have the height and the base, we can plug these values into the formula for the area of a right triangle:

\(\displaystyle A=\frac{1}{2}bh=\frac{1}{2}(12)(5)=\frac{1}{2}(60)=30\)

To find the area, use the following formula:

\(\displaystyle A=\frac{1}{2}Bh=\frac{1}{2}(4)(8)=16\)

In order for two triangles to be congruent, they must be identical. That is, the lengths of the corresponding sides of two congruent triangles must be equal. This means that in order for a triangle to be congruent to one with a height of  \(\displaystyle 5\)  and a base of  \(\displaystyle 12\),  its hypotenuse must be the same length as the hypotenuse of that triangle, which we can find using the Pythagorean Theorem:

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 5^2+12^2=c^2\)

\(\displaystyle c^2=169\rightarrow c=\sqrt{169}=13\)

In order for two right triangles to be congruent, the bases and heights must have identical lengths. Since we have a given right triangle with a base of \(\displaystyle 9\), the congruent right triangle must also have a base of \(\displaystyle 9\) .

In order for two right triangles to be congruent, the hypotenuses and acute angles must be identical. Since we have a given right triangle with an acute angle of \(\displaystyle 42^{\circ}\), the congruent right triangle must also have an acute angle of \(\displaystyle 42^{\circ}\) .

In order for two right triangles to be congruent, the bases and heights must have identical lengths. Since we have a given right triangle with a base of \(\displaystyle 12\), the congruent right triangle must also have a base of \(\displaystyle 12\) .

The hypotenuse of the large right triangle is 

\(\displaystyle \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25\)

The area of the large right triangle is half the product of its base and its height. The base can be any side of the triangle; the height would be the length of the altitude, which is the perpendicular segment from the opposite vertex to that base.

Therefore, the area of the triangle can be calculated as half the product of the legs:

\(\displaystyle A = \frac{1}{2} \cdot 7 \cdot 24 = 84\)

Or half the product of the hypotenuse and the length \(\displaystyle x\) of the dashed line.

\(\displaystyle A = \frac{1}{2} \cdot 25 \cdot x= \frac{25}{2} x\)

To calculate \(\displaystyle x\), we can set these expressions equal to each other:

\(\displaystyle \frac{25}{2} x = 84\)

\(\displaystyle \frac{25}{2} x \cdot \frac{2}{25}= 84\cdot \frac{2}{25}\)

\(\displaystyle x=6.72\)

Using the formula for the area of a right triangle, we can plug in the given values and solve for the height of the triangle:

\(\displaystyle A=\frac{1}{2}bh\)

\(\displaystyle 24=\frac{1}{2}(8)h\)

\(\displaystyle 24=4h\)

\(\displaystyle h=6\)

The height AE, divides the triangle ABC, in two triangles AEC and AEB with same proportions as the original triangle ABC, this property holds true for any right triangle.

In other words, \(\displaystyle \frac{AE}{AB}=\frac{AC}{BC}\).

Therefore, we can calculate, the length of AE: 

\(\displaystyle AE= \frac{AC\cdot AB}{BC}= \frac{12}{5}\).